1. **State the problem:** Find the derivative of the composite function $$y = \frac{1}{4} \ln \left( \frac{x - 1}{3x + 1} \right)$$ by changing the variable. 2. **Recall the formula:** The derivative of $$\ln u$$ with respect to $$x$$ is $$\frac{1}{u} \cdot \frac{du}{dx}$$. 3. **Set the inner function:** Let $$u = \frac{x - 1}{3x + 1}$$. 4. **Find $$\frac{du}{dx}$$ using the quotient rule:** $$\frac{du}{dx} = \frac{(3x + 1) \cdot \frac{d}{dx}(x - 1) - (x - 1) \cdot \frac{d}{dx}(3x + 1)}{(3x + 1)^2}$$ 5. **Calculate derivatives of numerator and denominator parts:** $$\frac{d}{dx}(x - 1) = 1$$ $$\frac{d}{dx}(3x + 1) = 3$$ 6. **Substitute these into the quotient rule:** $$\frac{du}{dx} = \frac{(3x + 1) \cdot 1 - (x - 1) \cdot 3}{(3x + 1)^2} = \frac{3x + 1 - 3x + 3}{(3x + 1)^2} = \frac{4}{(3x + 1)^2}$$ 7. **Now find $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} = \frac{1}{4} \cdot \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{4} \cdot \frac{1}{\frac{x - 1}{3x + 1}} \cdot \frac{4}{(3x + 1)^2}$$ 8. **Simplify the expression:** $$\frac{dy}{dx} = \frac{1}{4} \cdot \frac{3x + 1}{x - 1} \cdot \frac{4}{(3x + 1)^2} = \frac{\cancel{4}}{4} \cdot \frac{3x + 1}{x - 1} \cdot \frac{\cancel{4}}{(3x + 1)^2} = \frac{1}{1} \cdot \frac{3x + 1}{x - 1} \cdot \frac{1}{3x + 1} = \frac{1}{x - 1} \cdot \frac{1}{3x + 1} = \frac{1}{(x - 1)(3x + 1)}$$ **Final answer:** $$\boxed{\frac{dy}{dx} = \frac{1}{(x - 1)(3x + 1)}}$$